Question

For a positive integer n, let pn denote the product of the digits of n, and sn denote the sum of the digits of n. the number of integers between 10 and 1000 for which pn + sn = n is

Answer 1

case 1 :- Let positive integer n is two digit number .e.g., n = xy = 10x + y
now , Pn + Sn = n
xy + x + y = 10x + y
xy = 9x => y= 9 and $x\in \{1,2,3,4,5,6,7,8,9\}$
hence, pairs of (x,y) = (1,9), (2, 9) , (3,9) , (4,9) , (5,9), (6,9) , (7,9) , (8,9) , (9,9)
therefore 9 pairs of (x,y) between 10 to 100 where Pn + Sn = n .


case 2 :- Let n is a three digit number e.g., n = = xyz = 100x + 10y + z
now, Pn + Sn = n
=> xyz + x + y + z =100x + 10y + z
=> xyz = 99x + 9y
if we take x = 1 , and y = 1 then, z = 99 + 9 = 100
but x ,y , z should be 1 to 9 .
hence, there is no solution from here .


hence only 9 number of integer between 10 to 1000 where Pn + Sn = n